Sum HDU 4704

Sum
Time Limit: 1000msMemory Limit: 131072KBThis problem will be judged on HDU. Original ID: 4704
64-bit integer IO format: %I64d Java class name: Main
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Sample Input
2
Sample Output
2

这个就是找一个数的划分,比如 3有以下几种情况
3
1 2
2 1
1 1 1

规律求2 ^(n-1)
然后介绍 费马小定理
a的p次方,p为素数,且a,p互质,就有 a^(p-1)=p;

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>typedef long long ll;using namespace std;const ll N=(ll)1e9+7;ll quick_sort(ll a,ll b){    ll s=1;    while(b)    {        if(b&1)            s=s*a%N;        a=a*a%N;        b=b>>1;    }    return s%N;}char str[(int)1e5+10];int main(){  //  printf("%lld\n",N);    while(~scanf("%s",str))    {        ll num=0,ans=0;        for(int i=0;i<strlen(str);i++)//大数取模            num=(num*10+(str[i]-'0'))%(N-1);        if(!num)        {         //   puts("1");            ans=quick_sort(2,N-2);        }        else            ans=quick_sort(2,num-1);        printf("%lld\n",ans);    }    return 0;}